# An ansatz approach to the zeta function

** Published:**

The evaluation of Riemann’s zeta function

\[\zeta(s) = \sum_{k \geq 1} \frac{1}{k^s}\]at even integers is well-known, going back to the venerable Leonhard Euler and his answer to the Basel problem, namely

\[\zeta(2) = \frac{\pi^2}{6}.\]The Basel problem stumped even the great Bernoulli’s, and Euler’s solution
required the ingenuity and insight found throughout his work. Nearly
three-hundred years later, is this easier to answer? I claim that not only is
it easy, but that it is *entirely routine* to both *guess* the answer and to
*prove* it, provided that you allow some basic Fourier analysis. Let’s see how.

Given a suitably nice function $f$ on $[0, 1]$, we define its *Fourier
transform* by

These are the coefficients of the Fourier series

\[\sum_k \hat{f}(k) e^{2\pi i k}.\]There is a deep theory about what functions are suitably nice and what properties the Fourier transform satisfies, but for now we are only interested in one fact:

The family $\{e^{2\pi i n x}\}_n = \{e(n)\}_n$ is an orthonormal basis for the square-integrable functions on $[0, 1]$ equipped with the usual integral inner product

\[(f, g) = \int_0^1 f \overline{g}.\]In particular,

\[||f||_2 = \sum_k |(f, e(n))|^2 = \sum_k |\hat{f}(k)|^2.\]

In one direction, this tells us that sums whose terms are Fourier coefficients can be evaluated as an integral. This is exactly what we will do.

First, we need a humanly-proved lemma (though in principle a computer could likely figure this out).

**Lemma.** *The Fourier transform of $f_n(x) = x^n$ at integers satisfies
$\hat{f_n}(0) = (n + 1)^{-1}$, and $\hat{f}_n(k)$ is a polynomial in $(2\pi
i k)^{-1}$ of degree $n$ for all nonzero integers $k$.*

**Proof.** Apply integration by parts to prove that the sequence $\hat{f}_n(k)$
satisfies

The claim follows immediately by induction once we note that $\hat{f}_0(k) = 0$. $\blacksquare$

At this point we are in possession of a very powerful *ansatz*. The Fourier
transform of polynomials at integers gives us reciprocals of powers of
integers! There must be a connection with the zeta function. In particular, we
want to find a polynomial

such that $\hat{f}(k)$ is some multiple of $k^{-n}$ times a factor independent of $k$. Given the lemma above, we know that $\hat{f}(k)$, for nonzero $k$, is a polynomial in $(2\pi i)^{-1}$, meaning that we can look at their coefficients and require ones smaller than $(2\pi i)^{-n}$ to vanish. We can stipulate that $a_n = -1$ and that $\hat{f}(0) = 0$, which gives us a total of $n + 1$ linear equations in $n + 1$ unknowns, which we can (probably) solve!

So, in light of this, it suffices to write a program that will equate these
coefficients and solve the resulting linear equations. This will then, *a
posteriori*, provide evaluations of $\zeta(n)$ for positive, even integers $n$.
(Only the evens since, of course, we must take squares.)

# The program

Here is one such program that will do the job.

```
import sympy as sp
from sympy.abc import x, t
from sympy import I, pi
def zeta_ansatz(n):
xs = sp.symbols("a:{}".format(n + 1))
f = sum(xs[k] * x**k for k in range(n + 1))
k = sp.symbols("k", integer=True, zero=False)
ft = sp.integrate(f * sp.exp(-2 * pi * I * k * x), (x, 0, 1))
ft = ft.simplify().expand()
# Replace 2 pi I with a dummy variable 1 / t to grab coefficients.
ft = ft.subs(k, 1).subs(I, 1).subs(pi, 1 / t)
coeffs = ft.collect(t, evaluate=False)
zero_eqns = [coeff for key, coeff in coeffs.items() if key != t**n]
eqns = zero_eqns + [xs[-1] + 1, sum(x / (k + 1) for k, x in enumerate(xs))]
soln = sp.solve(eqns, xs)
f = sum(soln[xs[k]] * x**k for k in range(n + 1))
coeff = sp.factorial(n) / (2 * pi)**n
return f, sp.integrate(abs(f)**2, (x, 0, 1)) / coeff**2 / 2
```

An example:

```
In [1]: time [zeta_ansatz(k) for k in range(1, 5)]
CPU times: user 1.85 s, sys: 19 µs, total: 1.85 s
Wall time: 1.85 s
Out[1]:
⎡⎛ 2⎞ ⎛ 4⎞ ⎛ 2 6⎞ ⎛ 8 ⎞⎤
⎢⎜ π ⎟ ⎜ 2 1 π ⎟ ⎜ 3 3⋅x x π ⎟ ⎜ 4 3 2 1 π ⎟⎥
⎢⎜1/2 - x, ──⎟, ⎜- x + x - ─, ──⎟, ⎜- x + ──── - ─, ───⎟, ⎜- x + 2⋅x - x + ──, ────⎟⎥
⎣⎝ 6 ⎠ ⎝ 6 90⎠ ⎝ 2 2 945⎠ ⎝ 30 9450⎠⎦
```

So there, we have answered the Basel problem plus evaluated the next three terms of the sequence $\zeta(2n)$, all in under two seconds and without any foreknowledge of the answer. Not too shabby, eh, Euler?

# Connections to known results

These results are obviously not new. There are known closed-form evaluations of $\zeta(2n)$ for all positive integers $n$. A cursory glance suggests that the polynomials we get are exactly the negatives of the Bernoulli polynomials $B_n(x)$, and mentioned in that article is that the Fourier transform of $B_n(x)$ is

\[\hat{B}_n(x) = -\frac{n!}{(2\pi i)^n} \sum_{k \neq 0} \frac{e^{2\pi i k x}}{k^n},\]which matches exactly what we have said here.

Does this idea uniquely define the Bernoulli polynomials? The Fourier transform
is invertible, so saying “let $B_n(x)$ be the preimage of such and such
function under the Fourier transform on $[0, 1]$” is a fine definition. It is
surprising that such a thing is a *polynomial*, but nevertheless true, and
lucky for us that it was. Conversely, I am pretty sure that polynomials will
only ever give you linear combinations of the $\zeta$ function evaluated at
even integers, so we have completely exhausted the usefulness of polynomials
coupled with Fourier transforms.